# Find the range of the function $y = \sqrt{x^2 + 1}− x$?

I have a function $y = \sqrt{x^2 + 1} − x$, where the Domain is $(−\infty,+\infty)$.

Explanation for the domain

I need to make sure the domain of the function does not include values of $x$ that will make the square root negative.

This means that I need:

$x^2+1\ge0$

the discriminant is negative and $x \in\mathbb R$. In fact, $x^2\ge 0$, $\forall x\in\mathbb R$ and $1 > 0$.

I wish, if it's possible, to explain the value of the range with an algebraic demonstration. I am using an example to solve the function for $x$ using $y$ as parameter:

$\sqrt{x^2 + 1} − x = y$

$\sqrt{x^2 + 1} = y + x$

irrational equation, therefore:

$$\begin{cases} & y + x >= 0\\ & x^2 + 1 =(y+x)^2 \end{cases}$$

$$\begin{cases} & x >= -y\\ & x^2 + 1 =y^2 + x^2 + 2yx \end{cases}$$

$$\begin{cases} & x >= -y\\ & 1 =y^2 + 2yx \end{cases}$$

$$\begin{cases} & x >= -y\\ & x = \frac{1 - y^2}{2y} \end{cases}$$

now I need to find for which values is $x >= -y$, therefore:

$\frac{1 - y^2}{2y} < 0$

solving the numerator and denominator

numerator

$1 - y^2 > 0$

$y^2 -1 < 0$

solve for $-1 < y < +1$

denominator

$2y > 0$

solve for $y > 0$

combine the tow solutions the inequality has occurred when

$-1 < y < 0$ or $y> 1$

• Easier: $x^2$ is never negative, $x^2 + 1$ is even less negative. So, what is under $\sqrt\$ is never negative. – JnxF May 9 '16 at 22:26
• Thanks @JnxF is there a algebraic method to explain this conclusion? thanks – Gianni May 9 '16 at 22:28
• For all $x\in\mathbb{R}$ we have that $0\leq x^2 < x^2 +1$. – Eff May 9 '16 at 22:31
• $x^2 \geq 0 \implies x^2 + 1 \geq 1 \geq 0$. That's it. – JnxF May 9 '16 at 22:31
• @JnxF thanks but it's still not clear sorry. I find a method to revolve the function equation for $x$ and $y$ as a parameter. I am trying to use this example but without solution. – Gianni May 9 '16 at 22:36

Well, there is probably an easy algebraic method: $$y=\sqrt{x^2+1}-x$$$$\implies y^2+x^2+2xy=x^2+1$$$$\implies y^2+2xy-1=0$$ to find range of $y$, we reform this equation as $$0x^2+2xy+y^2-1=0$$ this is quadratic in $x$ and since $x$ is real , $y$ should also be real so discriminent should be greater than zero $$D=(2y)^2-4(y^2-1)(0)>0$$

and this implies $$y^2>0$$$$\implies y>0$$or$$y<0$$ but $y<0$ is ruled out since $|x|=\sqrt{x^2}<\sqrt{x^2+1}$, so $x<\sqrt{x^2+1}$ ,therefore $y>0$ for all values of $x$ so $$y\in (0,\infty)$$

• I got idea from OP – user5954246 May 9 '16 at 23:13

To determine the range you must study the variations of $y$ and apply the Intermediate value theorem (it's a continuous function).

Now $\;y'=\dfrac x{\sqrt{x^2+1}}-1=\dfrac {x-\sqrt{x^2+1}}{\sqrt{x^2+1}}<0,\;$ since $\;\lvert x\rvert<\sqrt{x^2+1}$. Hence the function is decreasing.

On the other hand,

• $\lim_{x\to-\infty}\sqrt{x^2+1}-x=+\infty+\infty=+\infty$,
• $\lim_{x\to+\infty}\sqrt{x^2+1}-x=\lim_{x\to+\infty}\dfrac{(x^2+1)-x^2}{\sqrt{x^2+1}+x}=0$.

Hence the range of $y$ is $(0,+\infty)$.

A purely computational solution:

As noted in another answer, $y=\dfrac1{x+\sqrt{x^2+1}}$. Hence the range is contained in $(0,+\infty)$. Lets show for any $m>0$, the equation $\dfrac1{x+\sqrt{x^2+1}}=m$ has a solution. Indeed, it is equivalent to $$\sqrt{x^2+1}=\frac1m-x\iff x^2+1=\Bigl(\frac1m-x\Bigr)^2\enspace\textbf{and}\enspace x \le\frac 1m.$$ Now $$x^2+1=\Bigl(\frac1m-x\Bigr)^2=\frac1{m^2}-\frac 2mx+x^2\iff x=\frac{1-m^2}{2m}$$ and the condition $\;x\le\dfrac1m$ is satisfied iff $\;1-m^2\le 2$, i.e. $m^2\ge -1$.

• Thanks @Bernard. I am honest to say we are study to find the range of function without using the limit. (see please the example above, thanks) – Gianni May 9 '16 at 22:56
• Then it comes down to solving quadratic equations. – Bernard May 9 '16 at 23:07
• I posted an example but I find a strange range and probably I wrong the approach. – Gianni May 9 '16 at 23:10
• @Gianni: I've added another solution, without using limits – Bernard May 9 '16 at 23:32

Hint: Expand the fraction with $\sqrt{x^2+1}+x$ to see that $$y=\frac{1}{\sqrt{x^2+1}+x}$$

This is only a partial answer as it only establishes the lower bound for the range.

Multiply by the conjugate

$$y=(\sqrt{x^2+1}-x) \cdot \dfrac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}$$

$$y=\dfrac{1}{\sqrt{x^2+1}+x}$$

This function is clearly approaching $0$ as $x$ is approaching positive infinity. Also the function is never negative, since $$1>0$$

$$x^2+1>x^2$$

$$\sqrt{x^2+1}>x$$

Therefore we established a lower bound of $0$ for your range.

Let's see:

$$\frac1{\sqrt{x^2+1}+x}=\frac1{-x\sqrt{1+\frac1{x^2}}+x}=\frac{-\frac1x}{\sqrt{1+\frac1{x^2}}-1}$$

With l'Hospital now (we have a $\;\frac00\;$ indeterminate);

$$\lim_{x\to-\infty}\frac{-\frac1x}{\sqrt{1+\frac1{x^2}}-1}=\lim_{x\to-\infty}\frac{\frac1{x^2}}{-\frac1{x^3\sqrt{1+\frac1{x^2}}}}=\lim_{x\to-\infty}-x\sqrt{1+\frac1{x^2}}=+\infty$$

And now:

$$\frac1{\sqrt{x^2+1}+x}=\frac1{x\sqrt{1+\frac1{x^2}}+x}=\frac{\frac1x}{\sqrt{1+\frac1{x^2}}+1}\xrightarrow[x\to\infty]{}\frac02=0$$

So I'd say the range is $\;(0,\infty)\;$ as this is a continuous function everywhere.

Other way: calculate any of the two limits above, and then check that

$$\left(\sqrt{x^2+1}-x\right)'=\frac x{\sqrt{x^2+1}}-1\le0\;\;\;\forall\,x\in\Bbb R$$

so the function is monotonic decreasing. Now, if you take the limit when $\;x\to-\infty\;$ then you use this and the fact that the function's always non-negative, and if you took the limit when $\;x\to\infty\;$ then use this and the fact the function's never constant in any interval.